sm chapter4

4Motion in Two Dimensions

CHAPTER OUTLINE

4.1 The Position, Velocity, and Acceleration Vectors

4.2 Two-Dimensional Motion with Constant Acceleration

4.3 Projectile Motion4.4 Uniform Circular Motion4.5 Tangential and

Radial Acceleration4.6 Relative Velocity and

Relative Acceleration

ANSWERS TO QUESTIONS

*Q4.1 The car’s acceleration must have an inward component and a forward component: answer (f ). Another argument: Draw a fi nal velocity vector of two units west. Add to it a vector of one unit south. This represents subtracting the initial velocity from the fi nal velocity, on the way to fi nding the acceleration. The direction of the resultant is that of vector (f ).

Q4.2 No, you cannot determine the instantaneous velocity. Yes, you can determine the average velocity. The points could be widely separated. In this case, you can only determine the average velocity, which is

��

vavg t= ∆

∆x

65

Q4.3 (a) (b)

*Q4.4 (i) The 45° angle means that at point A the horizontal and vertical velocity components are equal. The horizontal velocity component is the same at A, B, and C. The vertical velocity component is zero at B and negative at C. The assembled answer is a = b = c = e > d = 0 > f(ii) The x-component of acceleration is everywhere zero and the y-component is everywhere – 9.8 m �s2. Then we have a = c = e = 0 > b = d = f.

Q4.5 A parabola results, because the originally forward velocity component stays constant and the rocket motor gives the spacecraft constant acceleration in a perpendicular direction.

Q4.6 (a) yes (b) no: the escaping jet exhaust exerts an extra force on the plane. (c) no (d) yes (e) no: the stone is only a few times more dense than water, so friction is a signifi cant force onthe stone. The answer is (a) and (d).

Q4.7 The projectile is in free fall. Its vertical component of acceleration is the downward acceleration of gravity. Its horizontal component of acceleration is zero.

Q4.8 (a) no (b) yes (c) yes (d) no. Answer: (b) and (c)

*Q4.9 The projectile on the moon is in fl ight for a time interval six times larger, with the same range of vertical speeds and with the same constant horizontal speed as on Earth. Then (i) its range is (d) six times larger and (ii) its maximum altitude is (d) six times larger. Apollo astronauts performed the experiment with golf balls.

13794_04_ch04_p065-092.indd 6513794_04_ch04_p065-092.indd 65 11/28/06 1:16:09 PM11/28/06 1:16:09 PM

Q4.10 (a) no. Its velocity is constant in magnitude and direction. (b) yes. The particle is continuously changing the direction of its velocity vector.

Q4.11 (a) straight ahead (b) either in a circle or straight ahead. The acceleration magnitude can be constant either with a nonzero or with a zero value.

*Q4.12 (i) a = v2 �r becomes 32 �3 = 3 times larger: answer (b). (ii) T = 2π r �v changes by a factor of 3 �3 = 1. The answer is (a).

Q4.13

The skater starts at the center of the eight, goes clockwise around the left circle and then counter-clockwise around the right circle.

*Q4.14 With radius half as large, speed should be smaller by a factor of 1 � 2, so that a = v2 �r can be the same. The answer is (d).

*Q4.15 The wrench will hit (b) at the base of the mast. If air resistance is a factor, it will hit slightly leeward of the base of the mast, displaced in the direction in which air is moving relative to the deck. If the boat is scudding before the wind, for example, the wrench’s impact point can be in front of the mast.

*Q4.16 Let the positive x direction be that of the girl’s motion. The x component of the velocity of the ball relative to the ground is +5 – 12 m �s = –7 m �s. The x-velocity of the ball relative to the girl is –7 – 8 m �s = –15 m �s. The relative speed of the ball is +15 m �s, answer (d).

66 Chapter 4

SOLUTIONS TO PROBLEMS

Section 4.1 The Position, Velocity, and Acceleration Vectors

P4.1 x m( )

−−−

( )−

−

0

3 000

1 270

4 270

3 600

0

1 270

2 330m m

y m

(a) Net displacement at tan 1= + ( )=

x y x2 2

4 87

y −

�R . km at S of W28 6. °

(b) Average speeds s m s

=( )( ) + ( )20 0 180 25 0 120. .m s m s s

s s s

( ) + ( )( )+ +

=30 0 60 0

180 120 60 02

. .

.33 3. m s

(c) Average velocitym

360 sm s alo= × =4 87 10

13 53.

. nng�R

FIG. P4.1


Motion in Two Dimensions 67

P4.2 (a) �r i j= + −( )1 0 4 00 4 90 28. ˆ . . ˆt t t

(b) �v i j= ( ) + − ( )⎡⎣ ⎤⎦18 0 4 00 9 80 2. ˆ . ˆm s . m s m s t

(c) �a j= −( )9 80. ˆm s2

(d) by substitution, �r i j3 00 54 0 32 1. . ˆ . ˆs m m( ) = ( ) − ( )

(e) �v i j3 00 18 0 25 4. . ˆ . ˆs m s m s( ) = ( ) − ( )

(f ) �a j3 00 9 80. . ˆs m s2( ) = −( )

P4.3 The sun projects onto the ground the x component of her velocity:

5 00 0 2 50. cos . .m s m s−( ) =60 °

P4.4 (a) From x t= −5 00. sinω , the x component of velocity is

vx

dx

dt

d

dtt t= = ⎛

⎝⎞⎠ −( ) = −5 00 5 0. sin cosω ω ω.0

and ad

ttx

x= =vd

+ .5 00 2ω ωsin

similarly, vy

d

dtt t= ⎛

⎝⎞⎠ −( ) = +4 0 5 0 0 5 00. . cos sin0 0 ω ω ω.

and ad

dtt ty = ⎛

⎝⎞⎠ ( ) =5 0 5 00 2. sin cos0 ω ω ω ω.

At t = 0, �v i j i j= − + = +( )5 00 0 5 00 0 5 00 0. . mω ω ωcos ˆ sin ˆ . ˆ ˆ ss

and �a i j i j= + = +( )5 00 0 5 0 0 0 5 002 2 2. .ω ω ωsin ˆ cos ˆ ˆ . ˆ0 m s2

(b) �r i j j i= + = ( ) + ( ) − −x tˆ ˆ . ˆ . sin ˆ cosy 4 00 5 00m m ω ω tt j( )

�v i j= ( ) − +⎡⎣ ⎤⎦5 00. cos ˆ sin ˆm ω ω ωt t

�a i j= ( ) +⎡⎣ ⎤⎦5 00 2. sin ˆ cos ˆm ω ω ωt t

(c) The object moves in a circle of radius 5.00 m centered at 0 4, .000 m( ) .

Section 4.2 Two-Dimensional Motion with Constant Acceleration

P4.5 �v i ji = +( )4 00 1 00. ˆ . ˆ m s and

�v i j20.0 m s( ) = −( )20. ˆ . ˆ0 5 00

(a) atxx= = − =∆

∆v 20 0 4 0

20 00 800

. .

..

0m s m s2 2

atyy= = − − = −

∆∆v 5 00 1 0

20 00

. .

.

0m s .300 m s2 2

continued on next page


68 Chapter 4

(b) θ = −⎛⎝

⎞⎠ − = +−tan

.

..1 0 00

0 80020 6 339

3° °= from x aaxis

(c) At t = 25.0 s its position is specifi ed by its coordinates and the direction of its motion is specifi ed by the direction angle of its velocity:

x x t a tf i xi x= + + = + ( ) + ( )v1

210 0 4 00 25 0

1

20 8002 . . . . 225 0 360

1

24 00 1 00 2

2

2

.

. .

( ) =

= + + = − +

m

y y t a tf i yi yv 55 01

20 300 25 0 2. . .( ) + −( )( ) = −

= +

72.7 m

v vxf xi xa t == + ( ) =

= + = − ( ) = −

4 0 8 25 24

1 0 3 25 6

.

. .

m s

v vyf yi ya t 5 sm

θ =⎛⎝⎜

⎞⎠⎟

= −⎛⎝

⎞⎠ = −− −tan tan

.

.1 1 6 50

24 01

v

vy

x

55 2. °

P4.6 (a) �

�v

ri j j= = ⎛

⎝⎞⎠ −( ) = −d

dt

d

dtt t3 00 6 00 12 02. ˆ . ˆ . ˆ mm s

��

av

j j= = ⎛⎝

⎞⎠ −( ) = −d

dt dtt

d12 0 12 0. ˆ . ˆ m s2

(b) by substitution,� �r i j v j= −( ) = −3 00 6 00 12 0. ˆ . ˆ ; . ˆm m s

*P4.7 (a) From � � � �a v v a= =∫ ∫d dt d dt

i

f

i

f/ , we have

Then �v i j j− = = =∫5 6 6

3 241 2

3 2

00

3m/sˆ ˆ/

ˆ//

t dtt

tt

t

// /ˆ ˆ ˆ2 3 25 4so +j j�v i= t

(b) From � � � �v r r v= =∫ ∫d dt d dt

i

f

i

f/ , we have

Then �r i j i j− = ( ) = +

⎛⎝

0 5 4 5 45 2

3 25 2

+ˆ ˆ ˆ/

ˆ//

t dt tt

⎜⎜⎞⎠⎟

= +∫00

5 25 1 6t

t

t tˆ . ˆ/i j

P4.8 �a j= 3 00. ˆ m s2;

�v ii = 5 00. ˆ m s;

�r i ji = +0 0ˆ ˆ

(a) � � � �r r v a i jf i it t t t= + + = +⎡

⎣⎢⎤⎦

1

25 00

1

23 002 2. ˆ . ˆ

⎥⎥ m

� � �v v a i jf i t t= + = +( )5 00 3 00. ˆ . ˆ m s

(b) t = 2 00. s,�r i j if = ( ) + ( )( ) = +5 00 2 00

1

23 00 2 00 10 0 62. . ˆ . . ˆ . ˆ .. ˆ00 j( ) m

so x f = 10 0. m , yf = 6 00. m

�v i j i jf = + ( ) = +( )5 00 3 00 2 00 5 00 6 00. ˆ . . ˆ . ˆ . ˆ m s

v ff f xf yf= = + = ( ) + ( ) =�v v v2 2 2 25 00 6 00 7 81. . . m s



Section 4.3 Projectile Motion

P4.9 (a) The mug leaves the counter horizontally with a velocity vxi (say). If time t elapses before it hits the ground, then since there is no horizontal acceleration, x tf xi= v , i.e.,

tx f

xi xi

= = ( )v v

1 40. m

In the same time it falls a distance of 0.860 m with acceleration downward of 9 80. m s2. Then

y y t a tf i i y= + +vy

1

22: 0 0 860

1

29 0

1 402

= + −( )⎛⎝⎜

⎞⎠⎟

. ..

m m sm28

vxi

Thus,

vxi =( )( )

=4 90 1 96

3 34. .

.m s m

0.860 mm s

2 2

(b) The vertical velocity component with which it hits the fl oor is

v vyf yi ya t= + = + −( )⎛⎝⎜

⎞⎠

0 9 801 40

..

m sm

3.34 m s2

⎟⎟ = −4 11. m s

Hence, the angle θ at which the mug strikes the fl oor is given by

θ =⎛

⎝⎜⎞

⎠⎟= −⎛

⎝⎞⎠ = −− −tan tan

.

..1 1 4 11

3 3450

v

vyf

xf

99°

P4.10 The mug is a projectile from just after leaving the counter until just before it reaches the fl oor. Taking the origin at the point where the mug leaves the bar, the coordinates of the mug at any time are

x t a t tf xi x xi= + = +v v1

202 and y t a t g tf yi y= + = −v

1

20

1

22 2

When the mug reaches the fl oor, y hf = − so

− = −h g t1

22

which gives the time of impact as

th

g= 2

(a) Since x df = when the mug reaches the fl oor, x tf xi= v becomes dh

gxi= v2

giving the initial velocity as

vxi dg

h=

2

FIG. P4.9



70 Chapter 4

(b) Just before impact, the x component of velocity is still

v vxf xi=

while the y component is

v vyf yi ya t gh

g= + = −0

2

Then the direction of motion just before impact is below the horizontal at an angle of

θ =⎛

⎝⎜

⎞

⎠⎟ =

⎛⎝⎜

⎞⎠⎟

=− −tan tan/

/t1 1 2

2

v

vyf

xf

g h g

d g haan− ⎛

⎝⎞⎠

1 2h

d

The answer for vxi

indicates that a larger measured value for d would imply larger takeoff speed in direct proportion. A tape measure lying on the f loor could be calibrated as a speedometer. A larger value for h would imply a smaller value for speed by an inverse proportionality to the square root of h. That is, if h were nine times larger, v

xi would be three times

smaller. The answer for θ shows that the impact velocity makes an angle with the horizontal whose tangent is just twice as large as that of the elevation angle α of the edge of the table as seen from the impact point.

P4.11 x t t

x

xi i i= =

= ( )( )( )v v cos

cos

θ

300 55 0 42 0m s . . s°

xx

y t gt t gt

y

yi i i

= ×

= − = −

=

7 23 10

1

2

1

2

3

3

2 2

.

sin

m

v v θ

000 55 0 42 01

29 80 42 0m s . . s m s2( )( )( ) − ( )sin . .° s m( ) = ×2 31 68 10.

FIG. P4.10

h

dα

θ



P4.12 (a) To identify the maximum height we let i be the launch point and f be the highest point:

v v

v

yf yi y f i

i i

a y y

g y

2 2

2 2

2

0 2 0

= + −( )= + −( ) −sin maxθ (( )

=yg

i imax

sin.

v 2 2

2

θ

To identify the range we let i be the launch and f be the impact point; where t is not zero:

y t a t

t g t

t

f i yi y

i i

i

= + +

= + + −( )

=

y v

v

v

1

2

0 01

22

2

2sinθ

ssin

cossin

θ

θ θ

i

f i xi x

i ii i

g

x x t a t

d

= + +

= +

v

vv

1

2

02

2

gg+ 0

For this rock, d y= max

v vi i i i i

i

ii

g g

2 2 2

2

2sin sin cos

sin

costan

θ θ θ

θθ

θ

=

= ==

=

4

76 0θi . °

(b) Since g divides out, the answer is the same on every planet .

(c) The maximum range is attained forθi = 45°:

d

d

g

gi i

i i

max cos sin

cos sin= =v v

v v45 2 45

76 2 76

° °

° °22 125.

So dd

max = 17

8.

P4.13 hg

i i= v2 2

2

sin θ; R

gi i=

( )v2 2sin θ; 3h R=

so 3

2

22 2 2v vi i i i

g g

sin sinθ θ=

( )

or 2

3 2 2

2

= =sin

sin

tanθθ

θi

i

i

thus θi = ⎛⎝

⎞⎠ =−tan .1 4

353 1°


72 Chapter 4

P4.14 The horizontal component of displacement is x t tf xi i i= = ( )v v cosθ . Therefore, the time required

to reach the building a distance d away is td

i i

=v cosθ

. At this time, the altitude of the water is

y t a td g d

f yi y i ii i i

= + =⎛⎝⎜

⎞⎠⎟

−v vv v

1

2 22 sin

cos cθ

θ oosθi

⎛⎝⎜

⎞⎠⎟

2

Therefore the water strikes the building at a height h above ground level of

h y dgd

f ii i

= = −tancos

θθ

2

2 22v

P4.15 (a) x tf xi= = ( ) =v 8 00 20 0 3 00 22 6. cos . . .° m

(b) Taking y positive downwards,

y t g t

y

f yi

f

= +

= ( ) + ( )

v1

2

8 00 20 0 3 001

29 80

2

. sin . . .° 33 00 52 32. .( ) = m

(c) 10 0 8 00 20 01

29 80 2. . sin . .= ( ) + ( )° t t

4 90 2 74 10 0 0

2 74 2 74 196

9 80

2

2

. . .

. .

.

t t

t

+ − =

=− ± ( ) +

= 11 18. s

*P4.16 The time of fl ight of a water drop is given by y y t a tf i yi y= + +v1

22.

0 2 35 0

1

29 8

12= + − ( ). .m m s2 t

For t1

0> , the root is t1

2 2 35

9 80 693= ( )

=.

..

m

m ss2 .

(a) The horizontal range of the font is

x x t a tf i xi x121

20 1 70 0 693 0 1 1

= + +

= + ( ) + =

v

. . .m s s 88 m

This is about the width of a town sidewalk, so there is space for a walkway behind the

waterfall. Unless the lip of the channel is well designed, water may drip on the visitors.

A tall or inattentive person may get his head wet.

(b) Now the fl ight time t2

is given by 0 01

22 22= + −y gt .

ty

g

y

g

y

g

t2

2 1 1 12 2

12

1

12

2

12= =

( ) = = From the same equation as in part (a) for

horizontal range, x t2 2 2= v .

x t1

21

12 12= v v

v2

1

1

1

12 12

1 70

120 491= = = =x

t

..

m sm s

The rule that the scale factor for speed is the square root of the scale factor for distance is Froude’s law, published in 1870.

FIG. P4.16

1.70 m/s

2.35 m



P4.17 (a) We use the trajectory equation:

y xgx

f f if

i i

= −tancos

θθ

2

2 22v

Withx f = 36 0. m , vi = 20 0. m s, and θ = 53 0. °

we fi nd

yf = ( ) −( )( )

36 0 53 09 80 36 0

2 20

2

. tan .. .

.m

m s m2

°00 53 0

3 942 2m sm

( ) ( )=

cos ..

°

The ball clears the bar by

3 94 3 05 0 889. . .−( ) =m m

(b) The time the ball takes to reach the maximum height is

tg

i i1

20 0 53 0

9 801 6= =

( )( ) =v sin . .

..

θ m s sin

m s2

°33 s

The time to travel 36.0 m horizontally is tx f

ix2 =

v

t2

36 0

20 0 53 02 99= ( ) =.

( . cos ..

m

m s)s

°

Since t t2 1

> the ball clears the goal on its way down .

P4.18 When the bomb has fallen a vertical distance 2.15 km, it has traveled a horizontal distance x f given by

x

y x

f

f f

= ( ) − ( ) =

= −

3 25 2 15 2 4372 2. . .

tan

km km km

θggx f

i i

i

2

2 22

2 150 2 4379 8

v cos

tan.

θ

θ− = ( ) −m mm s22 m

m s

m

( )( )( )

∴− =

2 437

2 280

2 150 2 437

2

2 2cos θi

mm m( ) − ( ) +( )∴ −

tan . tan

tan . t

θ θ

θi i371 19 1

6 565

2

2 aan .

tan . .

θ

θ

i

i

− =

∴ = ± ( ) − ( ) −

4 792 0

1

26 565 6 565 4 1 42 .. . .792 3 283 3 945( )( ) = ±

Select the negative solution, sinceθi is below the horizontal.

∴ = −tan .θi 0 662, θi = −33 5. °

P4.19 (a) For the horizontal motion, we have

x x t a tf i xi x

i

= + +

= + ( )( ) +

v

v

1

224 0 53 2 2 0

2

m scos .°

vvi = 18 1. .m s



74 Chapter 4

(b) As it passes over the wall, the ball is above the street by y y t a tf i yi y= + +v1

22

yf = + ( )( )( ) + −( )0 18 1 53 2 21

29 8 2. sin . .m s s m s2° .. .2 8 132s m( ) =

So it clears the parapet by 8 13 7 1 13. .m m m− = .

(c) Note that the highest point of the ball’s trajectory is not directly above the wall. For the whole fl ight, we have from the trajectory equation

y xg

xf i fi i

f= ( ) −⎛⎝⎜

⎞⎠⎟

tancos

θθ2 2 2

2

v

or

6 539 8

18 1 532 2

mm s

m s

2

= ( ) −( )

⎛

⎝⎜tan

.

. cos°

2 °x f

⎞⎞

⎠⎟x f

2

Solving,

0 041 2 1 33 6 01 2. .m m−( ) − + =x xf f

and

x f =± − ( )( )

( )−

1 33 1 33 4 0 0412 6

2 0 0412

2

1

. . .

. m

This yields two results:

x f = 26 8. m or 5.44 m

The ball passes twice through the level of the roof. It hits the roof at distance from the wall

26 8 24 2 79. .m m m− =

P4.20 From the instant he leaves the fl oor until just before he lands, the basketball star is a projectile.

His vertical velocity and vertical displacement are related by the equation v vyf yi y f ia y y2 2 2= + −( ). Applying this to the upward part of his fl ight gives 0 2 9 80 1 85 1 022= + −( ) −( )vyi . . .m s m2 . From this, vyi = 4 03. m s. [Note that this is the answer to part (c) of this problem.]

For the downward part of the fl ight, the equation gives vyf2 0 2 9 80 0 900 1 85= + −( ) −( ). . .m s m2 .

Thus the vertical velocity just before he lands is

vyf = −4 32. m s

(a) His hang time may then be found from v vyf yi ya t= + :

− = + −( )4 32 4 03 9 80. . .m s m s m s2 t

or t = 0 852. s .

(b) Looking at the total horizontal displacement during the leap, x txi= v becomes

2 80 0 852. .m s= ( )vxi

which yields vxi = 3 29. m s .

(c) vyi = 4.03 m s . See above for proof.




(d) The takeoff angle is:θ =⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞− −tan tan.1 1 4 03v

vyi

xi

m s

3.29 m s⎠⎠⎟= 50 8. ° .

(e) Similarly for the deer, the upward part of the fl ight gives v vyf yi y f ia y y2 2 2= + −( ):

0 2 9 80 2 50 1 202= + −( ) −( )vyi . . .m s m2

so vyi = 5 04. m s.

For the downward part, v vyf yi y f ia y y2 2 2= + −( ) yields

vyf2 0 2 9 80 0 700 2 50= + −( ) −( ). . .m s m2 and vyf = −5 94. m s

The hang time is then found as v vyf yi ya t= + : − = + −( )5 94 5 04 9 80. . .m s m s m s2 t and

t =1 12. s

P4.21 The horizontal kick gives zero vertical velocity to the rock. Then its time of fl ight follows from

y y t a t

t

f i yi y= + +

− = + + −( )

v1

2

40 0 0 01

29 80

2

. .m m s2 22

2 86t = . .s

The extra time 3 00 2 86 0 143. . .s s s− = is the time required for the sound she hears to travel straight back to the player. It covers distance

343 0 143 49 0 40 02 2m s s m m( ) = = + ( ). . .x

where x represents the horizontal distance the rock travels.

x t txi

xi

= = +

∴ = =

28 3 0

28 3

2 869 91

2.

.

..

m

m

sm s

v

v

*P4.22 We match the given equations

x tf = + ( )

= +

0 11 2 18 5

0 360 0 840 11 2

. cos .

. . .

m s

m m

°

m s m s2( ) − ( )sin . .18 51

29 80 2°t t

to the equations for the coordinates of the fi nal position of a projectile

x x t

y y t gt

f i xi

f i yi

= +

= + −

v

v1

22

For the equations to represent the same functions of time, all coeffi cients must agree: xi = 0, yi = 0 840. m, vxi = ( )11 2 18 5. cos .m s °, vyi = ( )11 2 18 5. sin .m s ° and g = 9 80. m s2 .

(a) Then the original position of the athlete’s center of mass is the point with coordinates

x yi i, , .( ) = ( )0 0 840 m . That is, his original position has position vector

�r i j= +0 0 840ˆ . ˆm .



76 Chapter 4

(b) His original velocity is�v i ji = ( ) + ( )11 2 18 5 11 2 18 5. cos . ˆ . sin . ˆm s m s° ° ==

11 2. m s at 18.5° above the x axis.

(c) From 4 90 3 55 0 48 02. . .m s m s m2 t t− − = we fi nd the time of fl ight, which must be

positive t =+ + ( ) − ( ) −( )3 55 3 55 4 4 9 0 48

2 4

2. . . .m s m s m s m2

...

90 842

m ss

2( ) = . Then

x f = ( ) ( ) =11 2 18 5 0 8425 8 94. cos . . .m s m°

(d)

The free-fall trajectory of the athlete is a section around the vertex of a parabola opening downward, everywhere close to horizontal and 48 cm lower on the landing side than on the takeoff side.

P4.23 For the smallest impact angle

θ =⎛

⎝⎜⎞

⎠⎟−tan 1 v

vyf

xf

we want to minimize vyf and maximize v vxf xi= . The fi nal y component of velocity is related to vyi by v vyf yi gh2 2 2= + , so we want to minimize vyi and maximize vxi . Both are accomplished by making the initial velocity

horizontal. Then v vxi = , vyi = 0, and vyf gh= 2 . At last, the impact angle is

θ =⎛

⎝⎜⎞

⎠⎟=

⎛⎝⎜

⎞⎠⎟

− −tan tan1 1 2v

v vyf

xf

gh

Section 4.4 Uniform Circular Motion

P4.24 aR

= v2

, T = ( ) =24 3 600 86 400h s sh

v = = × =2 2 6 37 10463

6π πR

T

( . m)

86 400 sm s

a =( )

×=

463

6 37 100 033 76

2m s

mm s directed t

2

.. ooward the center of Earth

P4.25 ac = =( ) =v2 220 0

1 06377

r

.

.

m s

mm s2

The mass is unnecessary information.

FIG. P4.23

0.84 m 0.36 m

8.94 m

FIG. P4.22



P4.26 arc = v2

v = = ( )( ) =a rc 3 9 8 9 45 16 7. . .m s m m s2

Each revolution carries the astronaut over a distance of 2 2 9 45 59 4π πr = ( ) =. .m m. Then the rotation rate is

16 71

0 281. .m srev

59.4 mrev s

⎛⎝⎜

⎞⎠⎟

=

P4.27 (a) v = rω

At 8.00 rev s, v = ( )( )( )= =0 600 8 00 2 30 2 9. m . rev s rad rev . m sπ .. m s60π .

At 6.00 rev s, v = ( )( )( )= =0 900 6 00 2 33 9 1. m . rev s rad rev m sπ . 00 8. π m s.

6 00. rev s

gives the larger linear speed.

(b) Acceleration = =( ) = ×v2 2

39 60

0 6001 52 10

r

.

..

π m s

mm s2 .

(c) At 6.00 rev s, acceleration =( )

= ×10 8

0 9001 28 10

2

3.

..

π m s

mm s2 . So 8 rev/s gives

the higher acceleration.

Section 4.5 Tangential and Radial Acceleration

*P4.28 The particle’s centripetal acceleration is v2 �r = (3 m �s)2 �2 m = 4.50 m �s2. The total acceleration magnitude can be larger than or equal to this, but not smaller.

(a) Yes. The particle can be either speeding up or slowing down, with a tangential component

of acceleration of magnitude 6 4 5 3 972 2− =. . m/s2 .

(b) No. The magnitude of the acceleration cannot be less than v2 �r = 4.5 m �s2.

P4.29 We assume the train is still slowing down at the instant in question.

ar

at

c

t

= =

= =−( )( )

v

v

2

3

1 29

40 0 10 1

.

.

m s

km h m km

2

∆∆

hh s

sm s2/

..

.

3600

15 00 741

1 292 2

( )= −

= + =a a ac t mm s m s2 2( ) + −( )2 20 741.

at an angle of tan tan− −⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠

1 1a

at

c

0.741

1.29

�a = 1 48. m s inward and 29.9 backward2 °

FIG. P4.29


78 Chapter 4

P4.30 (a) See fi gure to the right.

(b) The components of the 20.2 and the 22 5. m s2

along the rope together constitute the centripetal acceleration:

ac = ( ) −( ) + ( )22 5 90 0 36 9 20 2. cos . . . cosm s m s2 2° ° 336 9 29 7. .° = m s2

(c) arc = v2

so v = = ( ) =a rc 29 7 1 50 6 67. . .m s m m s tangent to c2 iircle

�v = 6 67. m s at 36.9 above the horizontal°

P4.31 r = 2 50. m, a = 15 0. m s2

(a) ac = = ( )( ) =a cos . . cos .30 0 15 0 30 13 0° °m s m s2 2

(b) arc = v2

so v2 2 50 13 0 32 5= = ( ) =rac . . .m m s m s2 2 2

v = =32 5 5 70. .m s m s

(c) a a at r2 2 2= +

so a a at r= − = ( ) − ( ) =2 2 215 0 13 0 7 50. . .m s m s m s2 2 2

P4.32 Let i be the starting point and f be one revolution later. The curvilinear motion with constant tangential acceleration is described by

and v vxf xi xa t= + , v f ta tr

t= + =0

4π. The magnitude of the radial acceleration is a

r

r

t rrf= =

v2 2 2

2

16π.

Then tanθ ππ π

= = =a

a

r t

t rt

r

4

16

1

4

2

2 2 θ = 4 55. °

Section 4.6 Relative Velocity and Relative Acceleration

P4.33 �vce = the velocity of the car relative to the earth.

�vwc = the velocity of the water relative to the car.

�vwe = the velocity of the water relative to the earth.

These velocities are related as shown in the diagram at the right.

(a) Since �vwe is vertical, v vwc cesin . .60 0 50 0° = = km h or

�vwc = 57 7 60 0. .km h at west of vertical° .

(b) Since �vce has zero vertical component,

v vwe wc= = ( ) =cos . . cos . .60 0 57 7 60 0 28 9° °km h km h downward

∆ x t a t

r a t

ar

t

xi x

t

t

= +

= +

=

v1

2

2 01

24

2

2

2

π

π

FIG. P4.30

FIG. P4.31

FIG. P4.32

θa

at

ar

vce

60ºvwe vwc

we ce wc= +v v v

FIG. P4.33



P4.34 (a) � �

�v a i j

v

H H2

H

+ . . m s . s= = −( ) ( )

=

0 3 00 2 00 5 00

1

t ˆ ˆ

55 0 10 0

0 1 00 3 00

. . m s

+ . + .J

ˆ ˆ

ˆ ˆ

i j

v a i j

−( )= = (� �

jt )) ( )

= ( )=

m s . s

. + . m s

2

J

HJ H

5 00

5 00 15 0�

�v i j

v

ˆ ˆ

v −− = − − −( )=

vJ

HJ

. . . . m s15 0 10 0 5 00 15 0

1

ˆ ˆ ˆ ˆi j i j

v�

00 0 25 0

10 0 25 0 22 2

. . m s

m sHJ

ˆ ˆ

( . ) ( . )

i j

v

−( )= + =�

66 9. m s

(b) � �r a i jH H

2m s 5.00 s= + + = −( )0 01

2

1

23 00 2 002t . ˆ . ˆ (( )

= −( )= +

2

H

J

. . m�

�

r i j

r i

37 5 25 0

1

21 00 3 00

ˆ ˆ

. ˆ . jj i j

r r

( ) ( ) = +( )= −

m s 5.00s m2

HJ H

2 12 5 37 5. ˆ . ˆ

� � ��

�r i j i j

r

J

HJ

. . . . m= − − −( )=

37 5 25 0 12 5 37 5

25

ˆ ˆ ˆ ˆ

.. . m

mHJ

0 62 5

25 0 62 5 67 32 2

ˆ ˆ

. . .

i j

r

−( )= ( ) + ( ) =�

mm

(c) � � �a a a i j i jHJ H J . . . .= − = − − −( )3 00 2 00 1 00 3 00ˆ ˆ ˆ ˆ mm s

m s

2

HJ2�

a i j= −( )2 00 5 00. ˆ . ˆ

P4.35 Total time in still water td= = = ×v

2 000

1 201 67 103

.. s.

Total time = time upstream plus time downstream:

t

t

up

down

s=−

= ×

=

1000

1 20 0 5001 43 10

1000

1

3

( . . ).

.220 0 500588

+=

..s

Therefore, t total s= × + = ×1 43 10 588 2 02 103 3. . .

This is 12.0% larger than the time in still water.

P4.36 The bumpers are initially100 0 100m km= . apart. After time t the bumper of the leading car travels 40.0 t, while the bumper of the chasing car travels 60.0t. Since the cars are side by sideat time t, we have

0 100 40 0 60 0. . .+ =t t yielding

t = × =−5 00 10 18 03. .h s


80 Chapter 4

P4.37 To guess the answer, think of v just a little less than the speed c of the river. Then poor Alan will spend most of his time paddling upstream making very little progress. His time-averaged speed will be low and Beth will win the race.

Now we calculate: For Alan, his speed downstream is c + v, while his speed upstream is c − v . Therefore, the total time for Alan is

tL

cL

cL c

c1 2 2

2

1=

++

−=

−v v v/

/

For Beth, her cross-stream speed (both ways) is

c 2 2− v

Thus, the total time for Beth is tL

c

L c

c2 2 2 2 2

2 2

1=

−=

−v v

/

/.

Since 1 12

2− <v

c, t t

1 2> , or Beth, who swims cross-stream, returns fi rst.

*P4.38 We can fi nd the time of fl ight of the can by considering its horizontal motion:

16 m = (9.5 m �s) t + 0 t = 1.68 s

(a) For the boy to catch the can at the same location on the truck bed, he must throw it

straight up, at 0° to the vertical .

(b) For the free fall of the can, yf = y

i + v

yit + (1 �2)a

yt2:

0 = 0 + vyi (1.68 s) − (1 �2)(9.8 m �s2)(1.68 s)2 v

yi = 8.25 m �s

(c) The boy sees the can always over his head, traversing a straight line segment upward and then downward .

(d) The ground observer sees the can move as a projectile on a symmetric section of a parabola opening downward . Its initial velocity is

(9.52 + 8.252)1 �2 m �s = 12.6 m/s north at tan−1(8.25/9.5) = 41.0° above the horizontal

P4.39 Identify the student as the S′ observer and the professor as the S observer. For the initial motion in S′, we have

′′

= =v

vy

x

tan .60 0 3°

Let u represent the speed of S′ relative to S. Then because there is no x-motion in S, we can write v vx x u= ′ + = 0 so that ′ = − = −vx u 10 0. m s . Hence the ball is thrown backwards in S′. Then,

v v vy y x= ′ = ′ =3 10 0 3. m s

Using vy gh2 2= we fi nd

h =( )

( ) =10 0 3

2 9 8015 3

2.

..

m s

m sm2

The motion of the ball as seen by the student in S′ is shown in diagram (b). The view of the professor in S is shown in diagram (c).

FIG. P4.39



*P4.40 (a) To an observer at rest in the train car, the bolt accelerates downward and toward the rear of the train.

a = ( ) + ( ) =

=

2 50 9 80 10 1

2 50

2 2. . .

tan.

m s m s m s

m

2

θ ss

m s.

to the south from

2

29 800 255

14 3

.

.

=

=θ ° tthe vertical

To this observer, the bolt moves as if it were in a gravitational fi eld of 9.80 m �s2 down + 2.50 m �s2 south.

(b) a = 9 80. m s vertically downward2

(c) If it is at rest relative to the ceiling at release, the bolt moves on a straight line downward and southward at 14.3 degrees from the vertical.

(d) The bolt moves on a parabola with a vertical axis.

P4.41 Choose the x axis along the 20-km distance. The y components of the displacements of the ship and the speedboat must agree:

26 40 15 50

11

km h km h( ) −( ) = ( )

= −

t tsin sin

sin

° ° α

α 11 0

5012 7

.. .= °

The speedboat should head

15 12 7 27 7° ° °+ =. . east of north

Additional Problems

P4.42 (a) The speed at the top is v vx i i= = ( ) =cos cosθ 143 45 101m s m s° .

(b) In free fall the plane reaches altitude given by

v vyf yi y f ia y y2 2

2

2

0 143 45 2 9 8

= + −( )= ( ) + −m s sin .° m s ft

ft mf

2( ) −( )= +

y

y

f

f

31 000

31 000 5223 28. tt

mft

13 27 104⎛

⎝⎞⎠ = ×. .

(c) For the whole free fall motion v vyf yi ya t= +

− = + −( )=

101 101 9 8

20 6

m s m s m s

s

2.

.

t

t

(d) arc = v2

v = = ( ) =a rc 0 8 9 8 4 130 180. . ,m s m m s2

FIG. P4.41

15º

N

E

40º 25º

α

x

y


82 Chapter 4

*P4.43 (a) At every point in the trajectory, including the top, the acceleration is 9.80 m/s2 down .

(b) We fi rst fi nd the speed of the ball just before it hits the basket rim.

v xf 2 + v yf

2 = v xi 2 + v yi

2 + 2ay( y

f − y

i )

v f 2 = v i

2 + 2ay(y

f − y

i) = (10.6 m �s)2 + 2(−9.8 m �s2)(3.05 m − 0) = 52.6 m2 �s2

vf = 7.25 m �s. The ball’s rebound speed is v

yi = (7.25 m �s) �2 = 3.63 m �s

Now take the initial point just after the ball leaves the rim, and the fi nal point at the top of its bounce.

v yf 2 = v yi

2 + 2ay(y

f − y

i): 0 = (3.63 m �s)2 + 2(−9.8 m �s2)( y

f − 3.05 m)

yf = −(3.63 m �s)2 �2(−9.8 m �s2) + 3.05 m = 3.72 m

*P4.44 (a) Take the positive x axis pointing east. The ball is in free fall between the point just after it leaves the player’s hands, and the point just before it bonks the bird. Its horizontal compo-nent of velocity remains constant with the value

(10.6 m �s)cos 55° = 6.08 m �s

We need to know the time of fl ight up to the eagle. We consider the ball’s vertical motion:

vyf = v

yi + a

yt 0 = (10.6 m �s)sin 55° = (−9.8 m �s2)t t = −(8.68 m �s) �(−9.8 m �s2) = 0.886 s

The horizontal component of displacement from the player to the bird is

xf = x

i + v

xt = 0 + (6.08 m �s)(0.886 s) = 5.39 m

The downward fl ight takes the same time because the ball moves through the same verti-cal distance with the same range of vertical speeds, including zero vertical speed at one endpoint. The horizontal velocity component of the ball is −1.5(6.08 m �s) = −9.12 m �s.The fi nal horizontal coordinate of the ball is

xf = x

i + v

xt = 5.39 m + (−9.12 m/s)(0.886 s) = 5.39 m − 8.08 m/s = −2.69 m

The ball lands a distance of 2.69 m behind the player .

(b) The angle could be either positive or negative. Here is a conceptual argument: The hori-zontal bounce sends the ball 2.69 m behind the player. To shorten this distance, the bird wants to reduce the horizontal velocity component of the ball. It can do this either by send-ing the ball upward or downward relative to the horizontal.

Here is a mathematical argument: The height of the bird is (1 �2)(9.8 m �s2)(0.886 s)2 = 3.85 m.The ball’s fl ight from the bird to the player is described by the pair of equations

y y t a tf i yi y= + +v1

22 0 = 3.85 m + (9.12 m �s)(sin θ) t + (1 �2)( −9.8 m �s2)t2

and xf = x

i + v

xt 0 = 5.39 m + (−9.12 m �s)(cos θ) t

Eliminating t by substitution gives a quadratic equation in θ. This equation has two solutions.



P4.45 Refer to the sketch. We f ind it convenient to solve part (b)f irst.

(b) ∆ x txi= v ; substitution yields 130 35 0= ( )vi tcos . ° .

∆ y t atyi= +v1

22; substitution yields

20 0 35 0

1

29 80 2. sin . .= ( ) + −( )vi t t°

Solving the above by substituting vit = 159

gives 20 = 91 − 4.9 t2 so t = 3 81. s .

(a) substituting back gives vi = 41 7. m s

(c) v vyf i i gt= −sinθ , v vx i i= cosθ

At t = 3 81. s, vyf = − ( )( ) = −41 7 35 0 9 80 3 81 13 4. sin . . . .° m s

v

v v v

x

f x yf

= ( ) =

= + =

41 7 35 0 34 1

36 72 2

. cos . .

.

° m s

m ss .

P4.46 At any time t, the two drops have identical y-coordinates. The distance between the two drops is then just twice the magnitude of the horizontal displacement either drop has undergone. Therefore,

d x t t t txi i i i i= ( ) = ( ) = ( ) =2 2 2 2v v vcos cosθ θ

P4.47 (a) arc = =

( ) =v2 25 00

1 0025 0

.

..

m s

mm s2

a gt = = 9 80. m s2

(b) See fi gure to the right.

(c) a a ac t= + = ( ) + ( ) =2 2 2 225 0 9 80 26 8. . .m s m s m s2 2 2

φ =⎛⎝⎜

⎞⎠⎟

= =− −tan tan.

..1 1 9 80

25 021

a

at

c

m s

m s

2

2 44°

P4.48 (a) The moon’s gravitational acceleration is the probe’s centripetal acceleration: (For the moon’s radius, see end papers of text.)

ar

=

( ) =×

= ×

v

v

v

2

2

6

6

1

69 80

1 74 10

2 84 10

..

.

m sm

m

2

2 ss km s2 = 1 69.

(b) v = 2π r

T

Tr= = ×

×= ×2 2 1 74 10

6 47 106

3π πv

( ..

m)

1.69 10 m ss3 == 1 80. h

FIG. P4.45

FIG. P4.47


84 Chapter 4

*P4.49 (a) We fi nd the x coordinate from x = 12 t. We fi nd the y coordinate from 49 t − 4.9 t2. Then we fi nd the projectile’s distance from the origin as (x2 + y2)1/2, with these results:

t (s) 0 1 2 3 4 5 6 7 8 9 10 r (m) 0 45.7 82.0 109 127 136 138 133 124 117 120

(b) From the table, it looks like the magnitude of r is largest at a bit less than 6 s.The vector

�v tells how

�r is changing. If

�v at a particular point has a component along

�r, then �

r will be increasing in magnitude (if�v is at an angle less than 90° from

�r) or decreasing

(if the angle between�v and

�r is more than 90°). To be at a maximum, the distance from the

origin must be momentarily staying constant, and the only way this can happen is for the angle between velocity and displacement to be a right angle. Then

�r will be changing in

direction at that point, but not in magnitude.

(c) The requirement for perpendicularity can be defi ned as equality between the tangent of the angle between

�v and the x direction and the tangent of the angle between

�r and the y direc-

tion. In symbols this is (9.8t − 49) �12 = 12t �(49t − 4.9t2), which has the solution t = 5.70 s, giving in turn r = 138 m. Alternatively, we can require dr2 �dt = 0 = (d �dt)[(12t)2 + (49t − 4.9t2)2], which results in the same equation with the same solution.

*P4.50 (a) The time of fl ight must be positive. It is determined by yf = y

i + v

yit − (1 �2)a

yt2

0 = 1.2 + v0 sin 35° t – 4.9t2 from the quadratic formula as t =

+ +0 574 0 329 23 52

9 80 0

2. . .

.

v v

Then the range follows from x = vxit + 0 = v

0t as

x v v v v0 0 0

2020 164 3 0 002 299 0 047 94( )= + +. . . where x is in meters and v

0 is in

meters per second.

(b) Substituting v0 = 0.1 gives x v0( ) = 0.0410 m

(c) Substituting v0 = 100 gives x v0( ) =

961 m

(d) When v0 is small, v

0 2 becomes negligible. The expression x v0( )

simplifi es to

v v0 00 164 3 0 0 0 405. .+ + = Note that this gives nearly the answer to part (b).

(e) When v0 is large, v

0 is negligible in comparison to v

0 2 . Then x v0( )

simplifi es to

x v v v v0 0 0

2020 0 002 299 0 047 94 0 0959( )≈ + + =. . . v

02

This nearly gives the answer

to part (c).

(f ) The graph of x versus v0 starts from the origin as a straight line with slope 0.405 s. Then

it curves upward above this tangent line, getting closer and closer to the parabola x = (0.095 9 s2 �m) v

0 2



P4.51 The special conditions allowing use of the horizontal range equation applies. For the ball thrown at 45°,

D Rg

i= =45

2 90v sin

For the bouncing ball,

D R Rg g

i i= + = +( )

1 2

2 22 2 2v vsin / sinθ θ

where θ is the angle it makes with the ground when thrown and when bouncing.

(a) We require:

v v vi i i

g g g

2 2 22 2

4

24

5

26 6

= +

=

=

sin sin

sin

.

θ θ

θ

θ °

(b) The time for any symmetric parabolic fl ight is given by

y t gt

t gt

f yi

i i

= −

= −

v

v

1

2

01

2

2

2sin .θ

If t = 0 is the time the ball is thrown, then tg

i i= 2v sinθ

is the time at landing.

So for the ball thrown at 45.0°

tg

i45

2 45 0= v sin . °

For the bouncing ball,

t t tg g

i i i= + = +( )

=1 2

2 26 6 2 2 26 6 3v v vsin . / sin . si° ° nn .26 6°

g

The ratio of this time to that for no bounce is

3 26 6

2 45 0

1 34

1 410 949

vv

i

i

g

g

sin . /

sin . /

.

..

°

°= =

FIG. P4.51


86 Chapter 4

P4.52 Equation of bank:

Equations of moti

y x2 16 1= ( )oon: x t

y g t

i= ( )

= − ( )

v 2

1

232

Substitute for t from (2) into (3) y gx

i

= −⎛⎝⎜

⎞⎠⎟

1

2

2

2v. Equate y

from the bank equation to y from the equations of motion:

161

2 416

2

2

22 4

4

2

x gx g x

x xg

i i

= −⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ ⇒ − =

v vxx

i

3

4416 0

v−

⎛⎝⎜

⎞⎠⎟

=

From this, x = 0 or xg

i34

2

64= v and x =⎛⎝⎜

⎞⎠⎟

=410

9 8018 8

4

2

1 3

..

/

m . Also,

y gx

i

= −⎛⎝⎜

⎞⎠⎟

= − ( )( )( ) = −1

2

1

2

9 80 18 8

10 0

2

2

2

2v. .

.117 3. m

P4.53 (a) ∆ y g t= − 1

22 ; ∆ x ti= v

Combine the equations eliminating t:

∆ ∆y g

x

i

= −⎛⎝⎜

⎞⎠⎟

1

2

2

v

From this, ∆ ∆x

y

g i( ) = −⎛⎝⎜

⎞⎠⎟

2 22v

thus ∆ ∆x

y

gi= − = − − = × =v2

2752 3 000

9 806 80 10 6 803( )

.. . kkm .

(b) The plane has the same velocity as the bomb in the x direction. Therefore, the plane will be

3 000 m directly above the bomb when it hits the ground.

(c) When φ is measured from the vertical, tanφ = ∆∆

x

y

therefore, φ = ∆∆

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=− −tan tan .1 1 6 800

3 00066 2

x

y° .

FIG. P4.53

FIG. P4.52



P4.54 Measure heights above the level ground. The elevation yb of the ball follows

y R gtb = + −01

22

with x ti= v so y Rgx

bi

= −2

22v.

(a) The elevation yr of points on the rock is described by

y x Rr2 2 2+ =

We will have y yb r= at x = 0, but for all other x we require the ball to be above the rock surface as in y yb r> . Then y x Rb

2 2 2+ >

Rgx

x R

Rgx R g x

x

i

i i

−⎛⎝⎜

⎞⎠⎟

+ >

− + +

2

2

2

2 2

22

2

2 4

4

2

4

v

v v22 2

2 4

42

2

24

>

+ >

R

g xx

gx R

i iv v.

If this inequality is satisfi ed for x approaching zero, it will be true for all x. If the ball’s parabolic trajectory has large enough radius of curvature at the start, the ball will clear the

whole rock:1 2> gR

iv

vi gR>

(b) With vi gR= and yb = 0, we have 02

2

= −Rgx

gR or x R= 2 .

The distance from the rock’s base is

x R R− = −( )2 1

P4.55 (a) From Part (c), the raptor dives for 6 34 2 00 4 34. . . s− = undergoing displacement 197 m downward and

10 0 4 34 43 4. . . m( )( ) = forward.

v =

( ) + ( )=∆

∆d

t

197 43 4

4 3446 5

2 2.

.. m s

(b) α = −⎛⎝

⎞⎠ = −−tan

..1 197

43 477 6°

(c) 1971

22= gt , t = 6 34. s FIG. P4.55


88 Chapter 4

P4.56 (a) Coyote:

Roadrunner:

∆

∆

x at t= = ( )1

270 0

1

215 02 2; . .

xx t ti i= =v v; .70 0

Solving the above, we get

vi = 22 9. m s and t = 3 06. s

(b) At the edge of the cliff,

vxi at= = ( )( ) =15 0 3 06 45 8. . . m s

Substituting into ∆ y a ty= 1

22, we fi nd

− = −( )

=

= + =

1001

29 80

4 52

1

245 8

2

2

.

.

.

t

t

x t a txi x

s

∆ v (( )( ) + ( )( )4 521

215 0 4 52 2. . .s s

Solving, ∆ x = 360 m

(c) For the Coyote’s motion through the air

v v

v v

xf xi x

yf yi y

a t

a t

= + = + ( ) =

= +

45 8 15 4 52 114. . m s

== − ( ) = −0 9 80 4 52 44 3. . . m s

P4.57 (a) While on the incline

v v

v v

v

f i

f i

f

a x

at

2 2

2

2

0 2 4 00 50 0

20 0

− =

− =

− = ( )( )

∆

. .

. −− =

=

=

0 4 00

20 0

5 00

.

.

.

t

t

fv m s

s

(b) Initial free-fl ight conditions give us

vxi = =20 0 37 0 16 0. cos . .° m s

and

vyi = − = −20 0 37 0 12 0. sin . .° m s

v vxf xi= since ax = 0

v v

yf y yia y= − + = − −( ) −( )+ −( ) = −2 2 9 80 30 0 12 0

2 2∆ . . . 227 1. m s

v v v

f yfxf= + = ( ) + −( ) =2 2 2 2

16 0 27 1. . 31.5 m�s at 59.4º below the horizontal

(c) t1 5= s; ta

yf yi

y2

27 1 12 0

9 801 53=

−= − +

−=

v v . .

.. s

t t t= + =1 2 6 53. s

(d) ∆x txi= = ( ) =v 2 16 0 1 53 24 5. . . m

FIG. P4.57



P4.58 Think of shaking down the mercury in an old fever thermometer. Swing your hand through a circular arc, quickly reversing direction at the bottom end. Suppose your hand moves through one-quarter of a circle of radius 60 cm in 0.1 s. Its speed is

1

42 0 6

9π( )( )

≈. m

0.1 sm s

and its centripetal acceleration is v2

29

0 610

r≈ (

.~

m s)

mm s

22 .

The tangential acceleration of stopping and reversing the motion will make the total acceleration somewhat larger, but will not affect its order of magnitude.

P4.59 (a) ∆ x txi= v , ∆ y t gtyi= +v1

22

d tcos . . cos .50 0 10 0 15 0° °= ( )

and− = ( ) + −( )d t tsin . . sin . .50 0 10 0 15 0

1

29 80 2° °

Solving, d = 43 2. m and t = 2 88. s.

(b) Since ax = 0,

v v

v v

xf xi

yf yi ya t

= = =

= + =

10 0 15 0 9 66

10

. . m scos .

.

°

00 15 0 9 80 2 88 2sin . . .° − ( ) = − 5.6 m s

Air resistance would ordinarily decrease the values of the range and landing speed. As an airfoil, he can defl ect air downward so that the air defl ects him upward. This means he can get some lift and increase his distance.

P4.60 (a) The ice chest fl oats downstream 2 km in time t, so that 2 km = vwt. The upstream motion of the boat is described by d w= −( )v v 15 min. The downstream motion is described by

d tw+ = + −2 15km min)( )(v v . We eliminate tw

= 2 km

v and d by substitution:

v v v vv

−( ) + = +( ) −⎛⎝⎜

⎞⎠⎟w w

w

15 22

15min kmkm

min

vv vvv

v15 min 15 min km km km 15( ) − ( ) + = + −ww

2 2 2 min 15 min

30 min 2 km

km

( ) − ( )

( ) =

=

v

vvv

v

w

w

w

2

330 minkm h= 4 00. .

(b) In the reference frame of the water, the chest is motionless. The boat travels upstream for 15 min at speed v, and then downstream at the same speed, to return to the same point. Thus it travels for 30 min. During this time, the falls approach the chest at speed vw ,

traveling 2 km. Thus

vw

x

t= = =∆

∆2

4 00km

30 minkm h.

FIG. P4.59


90 Chapter 4

P4.61 Find the highest fi ring angle θH for which the projectile will clear the mountain peak; this will yield the range of the closest point of bombardment. Next fi nd the lowest fi ring angle; thiswill yield the maximum range under these conditions if both θH and θL are >45°; x = 2500 m,y = 1800m, vi = 250 m s.

y t gt t gt

x t

f yi i

f xi i

= − = ( ) −

= =

v v

v v

1

2

1

22 2sin

cos

θ

θ(( )t

Thus t

x f

i

=v cosθ

.

Substitute into the expression for yf

yx

gx

xf if

i

f

if= ( ) −

⎛⎝⎜

⎞⎠⎟

=vv v

sincos cos

taθθ θ

1

2

2

nncos

θθ

−gx f

i

2

2 22v

but 1

122

costan

θθ= + so y x

gxf f

f

i

= − +( )tan tanθ θ2

22

21

v and

02 2

2

22

2

2= − + +gx

xgx

yf

if

f

ifv v

tan tanθ θ

Substitute values, use the quadratic formula and fi nd

tan .θ = 3 905 or 1.197, which gives θH = 75 6. ° and θL = 50 1. °

Range at mθ θH

i H

g( ) = = ×v2

323 07 10

sin. from enemy ship

3 07 10 2 500 300 2703. × − − = m from shore

Range at mθ θL

i L

g( ) = = ×v2

326 28 10

sin. from enemy ship

6 28 10 2 500 300 3 48 103 3. .× − − = × from shore

Therefore, safe distance is < 270 m or > ×3 48 10. 3 m from the shore.

FIG. P4.61

P4.62 We follow the steps outlined in Example 4.7, eliminating td

i

= cos

cos

φθv

to fi nd

vv v

i

i i

d gdd

sin cos

cos

cos

cossin

θ φθ

φθ

φ− = −2 2

2 22 Clearing of fractions,

2 22 2 2 2v vi igdcos sin cos cos cos sinθ θ φ φ θ φ− = −

To maximize d as a function of θ, we differentiate through with respect to θ and setdd

dθ= 0:

2 22 2v vi i gdd

dcos cos cos sin sin cos cθ θ φ θ θ φ

θ+ −( ) − oos cos sin sin2 22 2φ θ θ φ= − −( )vi

We use the trigonometric identities from Appendix B4 cos cos sin2 2 2θ θ θ= − and

sin sin cos2 2θ θ θ= to fi nd cos cos sin sinφ θ θ φ2 2= . Next, sin

costan

φφ

φ= and cottan

21

2θ

θ=

give cot tan tan2 90 2φ φ θ= = −( )° so φ θ= −90 2° and θ φ= −452

° .



ANSWERS TO EVEN PROBLEMS

P4.2 (a) �r i j= + −( )18 0 4 00 4 90 2. ˆ . . ˆt t t (b)

�v i j= + −( )18 0 4 00 9 80. ˆ . ˆ. t (c)

�a j= −( )9 80. ˆm s2

(d) 54 0 32 1. ˆ . ˆm m( ) − ( )i j (e) 18 0 25 4. ˆ . ˆm s m s( ) − ( )i j (f ) −( )9 80. ˆm s2 j

P4.4 (a) �v i j= − +( )5 00 0. ˆ ˆω m s ;

�a i j= +( )0 5 00 2ˆ . ˆω m s2

(b) �r j= 4 00. ˆm + − −( )5 00. sin ˆ cos ˆm ω ωt ti j ;

�v i j= − +( )5 00. cos ˆ sin ˆm ω ω ωt t ;

�a i j= +( )5 00 2. sin ˆ cos ˆm ω ω ωt t

(c) a circle of radius 5.00 m centered at 0 4 00, . m( )

P4.6 (a) �v j= −12 0. ˆt m s;

�a j= −12 0. ˆ m s2 (b)

� �r i j v j= −( ) = −3 00 6 00 12 0. ˆ . ˆ ; . ˆm m s

P4.8 (a) �r i j= +( )5 00 1 50 2. ˆ . ˆt t m;

�v i j= +( )5 00 3 00. ˆ . t m s (b)

�r i j= +( )10 0 6 00. ˆ . ˆ m; 7 81. m s

P4.10 (a) dg

h2 horizontally (b) tan− ⎛

⎝⎞⎠

1 2h

d below the horizontal

P4.12 (a) 76.0° (b) the same on every planet. Mathematically, this is because the acceleration of

gravity divides out of the answer. (c) 17

8

d

P4.14 dgd

ii i

tancos

θθ

− ( )2

2 22v

P4.16 (a) Yes. (b) (1.70 m �s) � 12 = 0.491 m �s

P4.18 33.5° below the horizontal

P4.20 (a) 0.852 s; (b) 3 29. m s; (c) 4.03 m s;

(d) 50.8°; (e) 1.12 s

P4.22 (a)�ri = 0 i + 0.840 m j (b) 11.2 m �s at 18.5° (c) 8.94 m (d) The free-fall trajectory of the

athlete is a section around the vertex of a parabola opening downward, everywhere close to hori-zontal and 48 cm lower on the landing side than on the takeoff side.

P4.24 0 033 7 2. m s toward the center of the Earth

P4.26 0 281. rev s

P4.28 (a) Yes. The particle can be either speeding up or slowing down, with a tangential component of

acceleration of magnitude 6 4 5 3 972 2− =. . m/s2 . (b) No. The magnitude of the acceleration

cannot be less than v2 �r = 4.5 m �s2.

P4.30 (a) see the solution (b) 29 7. m s2 (c) 6 67. m s at 36.9° above the horizontal

P4.32 4 55. °

P4.34 (a) 26 9. m s (b) 67 3. m (c) 2 00 5 00. ˆ . ˆi j−( ) m s2

P4.36 18.0 s


92 Chapter 4

P4.38 (a) 0° (b) 8.25 m �s (c) The can traverses a straight line segment upward and then downward (d) A symmetric section of a parabola opening downward; 12.6 m �s north at 41.0° above the horizontal.

P4.40 (a) 10 1. m s2 at 14.3° south from the vertical (b) 9 80. m s2 vertically downward (c) The bolt moves on a parabola with its axis downward and tilting to the south. It lands south of the point directly below its starting point. (d) The bolt moves on a parabola with a vertical axis.

P4.42 (a) 101 m s (b) 3 27 104. × ft (c) 20 6. s (d) 180 m s

P4.44 (a) 2.69 m (b) The angle could be either positive or negative. The horizontal bounce sends the ball 2.69 m behind the player. To shorten this distance, the bird wants to reduce the horizontal velocity component of the ball. It can do this either by sending the ball upward or downward relative to the horizontal.

P4.46 2vi it cosθ

P4.48 (a)1 69. km s ; (b) 6 47 103. × s

P4.50 (a) x = v0(0.1643 + 0.002 299 v

0 2 )1 �2 + 0.047 98 v

0 2 where x is in meters and v

0 is in meters per

second, (b) 0.410 m (c) 961 m (d) x ≈ 0.405 v0 (e) x ≈ 0.095 9 v

0 2 (f ) The graph of x

versus v0 starts from the origin as a straight line with slope 0.405 s. Then it curves upward above

this tangent line, getting closer and closer to the parabola x = (0.095 9 s2 �m) v 0 2 .

P4.52 18 8 17 3. .m; m−( )

P4.54 (a) gR ; (b) 2 1−( ) R

P4.56 (a) 22.9 m �s (b) 360 m from the base of the cliff (c) �v = (114 i – 44.3 j ) m �s

P4.58 Imagine you have a sick child and are shaking down the mercury in an old fever thermometer. Starting with your hand at the level of your shoulder, move your hand down as fast as you can and snap it around an arc at the bottom. ~102 m �s2 ~ 10 g

P4.60 4.00 km � h

P4.62 see the solution


sm chapter4

Education